CBSE Class 12 Chemistry Question Paper 2025 Solved Code 56-1-3

SECTION A

Questions No. 1 to 16 are Multiple Choice type questions carrying 1 mark each.

1. Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be [1 Mark]
(A) +1.19 V
(B) +0.89 V
(C) +0.18 V
(D) +1.83 V

Answer: (B) +0.89 V

Teacher's Note:
a) The couple with the higher reduction potential (Sn⁴⁺/Sn²⁺) acts as the cathode.
b) \( E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.15 - (-0.74) = +0.89 \, V \).

2. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr³⁺ ion (Atomic no. : Cr = 24) is _________. [1 Mark]
(A) 2.87 B.M.
(B) 3.87 B.M.
(C) 3.47 B.M.
(D) 3.57 B.M.

Answer: (B) 3.87 B.M.

Teacher's Note:
a) Cr³⁺ is 3d³, so it has 3 unpaired electrons.
b) Use \( \mu = \sqrt{n(n+2)} = \sqrt{15} \approx 3.87 \) B.M.

3. In case of association, abnormal molar mass of solute will [1 Mark]
(A) increase
(B) decrease
(C) remain same
(D) first increase and then decrease

Answer: (A) increase

Teacher's Note:
a) Association reduces the number of particles, so the colligative property is lower than expected.
b) A lower colligative property gives a higher calculated molar mass (van't Hoff factor i less than 1).

4. Alkyl halides undergoing nucleophilic bimolecular substitution reaction involve [1 Mark]
(A) retention of configuration
(B) formation of racemic mixture
(C) inversion of configuration
(D) formation of carbocation

Answer: (C) inversion of configuration

Teacher's Note:
a) In S_N2, the nucleophile attacks from the back side, opposite to the leaving group.
b) Racemisation and carbocation formation belong to the S_N1 mechanism.

5. Arrange the following compounds in increasing order of their boiling points :
(i) (CH₃)₂CH - CH₂Br (ii) CH₃CH₂CH₂CH₂Br (iii) H₃C - C(CH₃)(Br) - CH₃
The correct order is [1 Mark]

[Figure: Structures of the three compounds. (i) A CH carbon carrying two CH₃ groups and a CH₂Br group. (ii) Straight chain CH₃CH₂CH₂CH₂Br. (iii) A central carbon bonded to three CH₃ groups (one above, one on each side) and a Br atom below.]

(A) \( (ii) \lt (i) \lt (iii) \)
(B) \( (i) \lt (ii) \lt (iii) \)
(C) \( (iii) \lt (i) \lt (ii) \)
(D) \( (iii) \lt (ii) \lt (i) \)

Answer: (C) \( (iii) \lt (i) \lt (ii) \)

Teacher's Note:
a) For isomeric haloalkanes, boiling point decreases as branching increases.
b) More branching gives a more spherical shape, smaller surface area and weaker van der Waals forces.

6. The correct IUPAC name of [Pt(NH₃)₂Cl₂]²⁺ is [1 Mark]
(A) Diamminedichloridoplatinum (II)
(B) Diamminedichloridoplatinum (IV)
(C) Diamminedichloridoplatinum (O)
(D) Diamminedichloridoplatinate (IV)

Answer: (B) Diamminedichloridoplatinum (IV)

Teacher's Note:
a) Oxidation state: x + 2(0) + 2(-1) = +2, so x = +4.
b) The suffix "-ate" is used only for anionic complexes; this is a cation, so "platinum" is used.

7. The acid formed when propyl magnesium bromide is treated with CO₂ followed by acid hydrolysis is : [1 Mark]
(A) C₃H₇COOH
(B) C₂H₅COOH
(C) CH₃COOH
(D) C₃H₇OH

Answer: (A) \( C_3H_7COOH \)

Teacher's Note:
a) Grignard reagent + CO₂ followed by hydrolysis gives a carboxylic acid with one extra carbon atom.
b) C₃H₇MgBr gives butanoic acid, C₃H₇COOH.

8. Acidified KMnO₄ oxidises sulphite to [1 Mark]
(A) S₂O₃^2-
(B) S₂O₈^2-
(C) SO₂(g)
(D) SO₄^2-

Answer: (D) \( SO_4^{2-} \)

Teacher's Note:
a) In acidic medium, sulphite (S in +4 state) is oxidised to sulphate (S in +6 state).
b) Equation: 2MnO₄^- + 5SO₃^2- + 6H⁺ → 2Mn²⁺ + 5SO₄^2- + 3H₂O.

9. Which is the correct order of acid strength from the following ? [1 Mark]
(A) \( C_6H_5OH \gt H_2O \gt ROH \)
(B) \( C_6H_5OH \gt ROH \gt H_2O \)
(C) \( ROH \gt C_6H_5OH \gt H_2O \)
(D) \( H_2O \gt C_6H_5OH \gt ROH \)

Answer: (A) \( C_6H_5OH \gt H_2O \gt ROH \)

Teacher's Note:
a) Phenoxide ion is stabilised by resonance, so phenol is the strongest acid here.
b) The alkyl group (+I effect) makes alcohols weaker acids than water.

10. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because _________. [1 Mark]
(A) it gains water due to osmosis
(B) it loses water due to reverse osmosis
(C) it gains water due to reverse osmosis
(D) it loses water due to osmosis

Answer: (D) it loses water due to osmosis

Teacher's Note:
a) The salt solution is hypertonic, so water moves out of the mango cells.
b) Reverse osmosis needs external pressure, which is not applied here.

11. The best reagent for converting propanamide into propanamine is _________. [1 Mark]
(A) excess H₂
(B) Br₂ in aqueous NaOH
(C) iodine in the presence of red phosphorus
(D) LiAlH₄ in ether

Answer: (D) \( LiAlH_4 \) in ether

Teacher's Note:
a) LiAlH₄ reduces CH₃CH₂CONH₂ to CH₃CH₂CH₂NH₂ with the same number of carbon atoms.
b) Br₂/NaOH (Hoffmann bromamide degradation) gives ethanamine, which has one carbon less.

12. Which of the following statements is not true about glucose ? [1 Mark]
(A) It is an aldohexose.
(B) On heating with HI it forms n-hexane.
(C) It exists in furanose form.
(D) It does not give Schiff's test.

Answer: (C) It exists in furanose form.

Teacher's Note:
a) Glucose exists in the six-membered pyranose form; fructose exists in the furanose form.
b) Glucose does not give Schiff's test, which shows the free -CHO group is absent in its cyclic form.

For questions number 13 to 16, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :

13. Assertion (A) : All naturally occurring \( \alpha \)-amino acids except glycine are optically active.
Reason (R) : Most naturally occurring amino acids have L-configuration. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) Amino acids other than glycine are optically active because the \( \alpha \)-carbon is chiral.
b) L-configuration is a fact about natural amino acids but does not explain optical activity.

14. Assertion (A) : The boiling point of ethanol is higher than that of methoxymethane.
Reason (R) : There is intramolecular hydrogen bonding in ethanol. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Ethanol has intermolecular (not intramolecular) hydrogen bonding, which raises its boiling point.
b) Watch for the words "inter" and "intra" in such reasons.

15. Assertion (A) : The boiling points of alkyl halides decrease in the order : \( RI \gt RBr \gt RCl \gt RF \).
Reason (R) : The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) The order RI, RBr, RCl, RF follows the decrease in size and mass of the halogen, which weakens van der Waals forces.
b) The Reason compares haloalkanes with hydrocarbons, so it does not explain the order among halides.

16. Assertion (A) : [Cr(H₂O)₆]Cl₂ and [Fe(H₂O)₆]Cl₂ are examples of homoleptic complexes.
Reason (R) : All the ligands attached to the metal are the same. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) A homoleptic complex has only one kind of ligand bound to the metal (here only H₂O).
b) Cl^- ions outside the square bracket are counter ions, not ligands.

SECTION B

17. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal ? Justify your answer. [2 Marks]

Answer:
1. Benzaldehyde is less reactive than propanal in nucleophilic addition reactions.
2. Due to resonance with the benzene ring, the polarity of the carbonyl group is reduced in benzaldehyde, so its carbonyl carbon is less electrophilic than the carbonyl carbon of propanal.

Teacher's Note:
a) One mark is for "less reactive" and one mark is for the reason.
b) The key words are "resonance" and "less electrophilic carbonyl carbon".

18. Complete and balance the following chemical equations : [2 Marks]
(a) 8MnO₄^- + 3S₂O₃^2- + H₂O →
(b) Cr₂O₇^2- + 3 Sn²⁺ + 14H⁺ →

Answer:
(a) 8MnO₄^- + 3S₂O₃^2- + H₂O → 8MnO₂ + 6SO₄^2- + 2OH^-
(b) Cr₂O₇^2- + 3Sn²⁺ + 14H⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Teacher's Note:
a) In neutral or faintly alkaline medium, MnO₄^- is reduced to MnO₂ and thiosulphate is oxidised to sulphate.
b) Always check that both atoms and charges are balanced on the two sides.

19. (A) Give reasons : [2 Marks]
(a) Cooking is faster in pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution ? What change in temperature would you observe after mixing liquids X and Y ?

Answer:
(a) Inside a pressure cooker the pressure is high, so the boiling point of water is higher. Food cooks at a higher temperature, so cooking is faster.
(b) The solution shows negative deviation from Raoult's law. The temperature increases on mixing (heat is released).

Teacher's Note:
a) Link higher pressure to higher boiling point; this link carries the mark.
b) Negative deviation means \( \Delta_{mix}V \lt 0 \) and \( \Delta_{mix}H \lt 0 \), so the solution warms up.

OR

(B) Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law ? Give an example. [2 Marks]

Answer:
1. Azeotropes are binary mixtures that have the same composition in the liquid and vapour phase and boil at a constant temperature.
2. Negative deviation forms a maximum boiling azeotrope, for example a mixture of 68% HNO₃ and 32% water by mass.

Teacher's Note:
a) Positive deviation gives a minimum boiling azeotrope (e.g. about 95% ethanol-water).
b) Any correct example earns the mark; the percentage may be omitted.

20. Identify A and B in each of the following reaction sequence : [2 Marks]
(a) CH₃CH₂Cl →(NaCN) A →(H₂/Ni) B
(b) C₆H₅NH₂ →(NaNO₂/HCl, 0 - 5 °C) A →(C₆H₅NH₂, H⁺) B

Answer:
(a) A = CH₃CH₂CN (propanenitrile); B = CH₃CH₂CH₂NH₂ (propan-1-amine)
(b) A = C₆H₅N₂⁺Cl^- (benzenediazonium chloride); B = C₆H₅-N=N-C₆H₄-NH₂ (p-aminoazobenzene, with the -NH₂ group para to the azo group)

Teacher's Note:
a) Each of A and B carries half a mark in both parts.
b) Diazonium salts couple with aniline at the para position to give a yellow azo dye.

21. What are the hydrolysis products of : [2 Marks]
(a) Sucrose
(b) Lactose

Answer:
(a) Sucrose gives D-(+)-glucose and D-(-)-fructose: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose)
(b) Lactose gives \( \beta \)-D-galactose and \( \beta \)-D-glucose: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (galactose) + C₆H₁₂O₆ (glucose)

Teacher's Note:
a) Naming both monosaccharides correctly earns the full mark for each part.
b) Do not mix up lactose (glucose + galactose) with maltose (glucose + glucose).

SECTION C

22. Henry's law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the number of moles of CO₂ in 500 ml of soda water when packed under 2.53 × 10⁵ Pa at the same temperature. [3 Marks]

Answer:
Given: \( K_H = 1.67 \times 10^{8} \, Pa \), \( p_{CO_2} = 2.53 \times 10^{5} \, Pa \), volume of water = 500 mL (about 500 g)
Henry's law: \( p_{CO_2} = K_H \, \chi_{CO_2} \)
\( \chi_{CO_2} = \dfrac{p_{CO_2}}{K_H} = \dfrac{2.53 \times 10^{5}}{1.67 \times 10^{8}} = 1.51 \times 10^{-3} \)
Moles of water: \( n_{H_2O} = \dfrac{500}{18} = 27.78 \, mol \)
Since \( n_{CO_2} \) is very small, \( \chi_{CO_2} \approx \dfrac{n_{CO_2}}{n_{H_2O}} \)
\( n_{CO_2} = 27.78 \times 1.51 \times 10^{-3} = 42.0 \times 10^{-3} \, mol \)
Number of moles of CO₂ = 0.042 mol

Teacher's Note:
a) Calculating the mole fraction correctly carries 1 mark; write the formula first.
b) Take 500 mL of water as 500 g (density 1 g/mL) to find moles of water.
c) Write the unit "mol" with the final answer.

23. Calculate \( \Delta_r G^{\circ} \) and log K_C of the reaction. [3 Marks]
2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
Given \( E^{\circ}_{Cr^{3+}/Cr} = -0.74 \, V \)
\( E^{\circ}_{Cd^{2+}/Cd} = -0.40 \, V \)
[R = 8.314 J K^-1 mol^-1, F = 96500 C mol^-1]

Answer:
Cr is oxidised (anode) and Cd²⁺ is reduced (cathode); number of electrons transferred, n = 6
\( E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = -0.40 - (-0.74) = +0.34 \, V \)
\( \Delta_r G^{\circ} = -nFE^{\circ}_{cell} \)
\( \Delta_r G^{\circ} = -(6 \times 96500 \times 0.34) \, J \, mol^{-1} = -196860 \, J \, mol^{-1} \) (= -196.86 kJ mol^-1)
\( \Delta_r G^{\circ} = -2.303 \, RT \log K_C \)
\( \log K_C = \dfrac{196860}{2.303 \times 8.314 \times 298} = \dfrac{196860}{5705.8} \)
\( \log K_C = 34.50 \)

Teacher's Note:
a) n = 6 because 2 Cr atoms each lose 3 electrons; using n = 2 or 3 is a common mistake.
b) Keep the negative sign in \( \Delta_r G^{\circ} \); a negative value shows the reaction is spontaneous.
c) Any other correct method (for example \( \log K_C = \dfrac{nE^{\circ}}{0.059} \)) is also accepted.

24. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. [3 Marks]
[Given : log 4 = 0.602, log 2 = 0.301, R = 8.314 J K^-1 mol^-1]

Answer:
Given: \( T_1 = 293 \, K \), \( T_2 = 313 \, K \), \( \dfrac{k_2}{k_1} = 4 \)
Arrhenius equation: \( \log \dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R} \left[ \dfrac{T_2 - T_1}{T_1 T_2} \right] \)
\( \log 4 = \dfrac{E_a}{2.303 \times 8.314} \left[ \dfrac{313 - 293}{313 \times 293} \right] \)
\( 0.602 = \dfrac{E_a}{19.147} \times \dfrac{20}{313 \times 293} \)
\( E_a = \dfrac{0.602 \times 19.147 \times 313 \times 293}{20} \)
\( E_a = 52850 \, J \, mol^{-1} = 52.85 \, kJ \, mol^{-1} \)

Teacher's Note:
a) Writing the correct Arrhenius equation carries 1 mark.
b) "Rate quadruples" means \( k_2 / k_1 = 4 \), so use log 4 = 0.602.
c) Give the unit J mol^-1 or kJ mol^-1 with the final answer.

25. (A) Draw the structure of the major monohalo product for each of the following reaction : [3 Marks]
(a) Cl-C₆H₄-CH₂-CH₃ →(Br₂, Heat) ?
(b) CH₃-cyclohexene + HBr → ?
(c) HO-H₂C-C₆H₄-OH →(HCl, Heat) ?

[Figure: (a) A benzene ring with a Cl atom at one position and a CH₂ - CH₃ group at the para position, reacting with Br₂, Heat. (b) A cyclohexene ring with a CH₃ group on one carbon of the double bond, reacting with HBr. (c) A benzene ring with an OH group and a HO - H₂C group at the para position, reacting with HCl, Heat.]

Answer:
(a) Bromination takes place at the benzylic carbon. Product: Cl-C₆H₄-CH(Br)-CH₃, i.e. 1-(1-bromoethyl)-4-chlorobenzene (Cl and the CHBrCH₃ group para to each other).
(b) HBr adds by Markovnikov's rule. Product: 1-bromo-1-methylcyclohexane (Br and CH₃ on the same ring carbon).
(c) Only the benzylic -CH₂OH group is replaced by Cl; the phenolic -OH does not react. Product: HO-C₆H₄-CH₂Cl, i.e. 4-(chloromethyl)phenol.

Teacher's Note:
a) Free radical bromination on heating prefers the benzylic position because the benzylic radical is resonance stabilised.
b) The C-O bond of phenol has partial double bond character, so phenolic -OH is not replaced by HCl.

OR

(B) How do you convert : [3 Marks]
(a) Chlorobenzene to biphenyl
(b) Propene to 1-Iodopropane
(c) 2-bromobutane to but-2-ene.

Answer:
(a) Fittig reaction: 2C₆H₅Cl + 2Na →(dry ether) C₆H₅-C₆H₅ + 2NaCl
(b) Anti-Markovnikov addition, then Finkelstein reaction:
CH₃CH=CH₂ + HBr →(peroxide) CH₃CH₂CH₂Br
CH₃CH₂CH₂Br + NaI →(dry acetone) CH₃CH₂CH₂I + NaBr
(c) Dehydrohalogenation: CH₃CH(Br)CH₂CH₃ + alc. KOH →(heat) CH₃CH=CHCH₃ (major, Saytzeff product) + KBr + H₂O

Teacher's Note:
a) Always write the reagent and condition (dry ether, peroxide, dry acetone, alcoholic KOH) over the arrow.
b) Peroxide is essential in (b); without it Markovnikov addition gives 2-bromopropane.
c) Saytzeff rule: the more substituted alkene (but-2-ene) is the major product.

26. The elements of 3d transition series are given as :
Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
Answer the following : [3 Marks]
(a) Copper has exceptionally positive \( E^{\circ}_{M^{2+}/M} \) value, why ?
(b) Which element is a strong reducing agent in +2 oxidation state and why ?
(c) Zn²⁺ salts are colourless. Why ?

Answer:
(a) Copper has a high enthalpy of atomisation \( (\Delta_a H^{\circ}) \) and a low enthalpy of hydration \( (\Delta_{hyd} H^{\circ}) \). The energy needed to form Cu²⁺(aq) is not balanced by its hydration energy, so its \( E^{\circ} \) is positive.
(b) Chromium (Cr). Cr²⁺ easily changes to Cr³⁺ (d⁴ to d³), which has a stable half-filled t_2g level.
(c) Zn²⁺ has completely filled d-orbitals (3d¹⁰) with no unpaired electron, so no d-d transition is possible and its salts are colourless.

Teacher's Note:
a) In (b), naming Cr earns half a mark and the d³ / half-filled t_2g reason earns the other half.
b) Colour in transition metal ions needs unpaired d electrons for d-d transitions.

27. (a) Arrange the following compounds in increasing order of their boiling point :
(CH₃)₂NH, CH₃CH₂NH₂, CH₃CH₂OH.
(b) Give plausible explanation for each of the following :
(i) Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis.
(ii) Amides are less basic than amines. [3 Marks]

Answer:
(a) \( (CH_3)_2NH \lt CH_3CH_2NH_2 \lt CH_3CH_2OH \)
(b) (i) Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide, so aromatic primary amines cannot be made by this method.
(ii) In amides, the lone pair on nitrogen is delocalised onto the carbonyl oxygen by resonance (-R effect of C=O). So the lone pair is less available for donation and amides are less basic than amines.

Teacher's Note:
a) Alcohols form stronger hydrogen bonds than amines because O is more electronegative than N.
b) Primary amines have two N-H bonds for hydrogen bonding, secondary amines only one, so the primary amine boils higher.
c) In (b)(ii), showing the resonance structures of the amide group also earns the mark.

28. Define the following terms : [3 Marks]
(a) Native protein
(b) Nucleotide
(c) Essential amino acid

Answer:
(a) Native protein: a protein found in a biological system with a unique three-dimensional structure and biological activity.
(b) Nucleotide: a unit made of a nitrogenous base, a pentose sugar and a phosphate group; it is formed when a nucleoside is linked to phosphoric acid.
(c) Essential amino acids: amino acids that cannot be synthesised in the body and must be obtained through diet (for example valine, leucine, lysine).

Teacher's Note:
a) Do not confuse a nucleoside (base + sugar) with a nucleotide (base + sugar + phosphate).
b) For native protein, the words "biological activity" are the key value point.

SECTION D

The following questions are case based questions. Read the passage carefully and answer the questions that follow.

29. The rate of a chemical reaction is expressed either in terms of decrease in the concentration of reactants or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of reactants, concentration of reactants, temperature, presence of catalyst, surface area of the reactants and presence of light. Rate of reaction is directly related to the concentration of reactant. Rate law states that the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the Rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction.
Answer the following questions :

(a) (i) What is a rate determining step ?
(ii) Define complex reaction. [2 Marks]

Answer:
(i) The slowest step in a reaction mechanism, which decides the overall rate of the reaction, is called the rate determining step.
(ii) A complex reaction is a reaction that takes place in a series of elementary reactions, that is, in two or more steps.

Teacher's Note:
a) The key word for (i) is "slowest step".
b) For (ii), mention "sequence of elementary steps".

(b) What is the effect of temperature on the rate constant of a reaction ? [1 Mark]

Answer: The rate constant increases with increase in temperature (it nearly doubles for a 10 K rise).

Teacher's Note:
a) This follows from the Arrhenius equation \( k = Ae^{-E_a/RT} \).
b) A higher temperature gives more molecules with energy equal to or more than \( E_a \).

OR

(b) Why is molecularity applicable only for elementary reactions whereas order is applicable for elementary as well as complex reactions ? [1 Mark]

Answer: Molecularity is defined only for a single elementary step; in a complex reaction each step may have a different molecularity, so it has no meaning for the overall reaction. Order is determined experimentally (by the slowest step), so it applies to both elementary and complex reactions.

Teacher's Note:
a) Molecularity is theoretical; order is experimental. This contrast is the key point.
b) Molecularity is always a whole number, while order can be zero or fractional.

(c) The conversion of molecule X to Y follows second order kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y ? [1 Mark]

Answer: Rate \( = k[X]^2 \). New rate \( = k(3[X])^2 = 9k[X]^2 \), so the rate of formation of Y becomes 9 times.

Teacher's Note:
a) For second order, the rate changes as the square of the concentration change.
b) Write the rate law first; it makes the answer clear.

30. Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of OH group attached to the benzene ring. Since, the OH group increases the electron density more to o- and p- positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols.
Answer the following questions :

(a) What happens when phenol reacts with
(i) Br₂/CS₂
(ii) Conc. HNO₃ [2 Marks]

Answer:
(i) With bromine in CS₂ at low temperature (273 K), phenol gives a mixture of 2-bromophenol and 4-bromophenol (monobromophenols).
C₆H₅OH + Br₂ →(CS₂, 273 K) o-BrC₆H₄OH + p-BrC₆H₄OH
(ii) With concentrated HNO₃, phenol gives 2,4,6-trinitrophenol (picric acid).
C₆H₅OH + 3HNO₃ (conc.) → C₆H₂(NO₂)₃OH + 3H₂O

Teacher's Note:
a) A non-polar solvent (CS₂) gives only monobromination; bromine water gives 2,4,6-tribromophenol.
b) Naming the product (2-bromophenol, 4-bromophenol, picric acid) is enough for the marks.

(b) Why phenol does not undergo protonation readily ? [1 Mark]

Answer: Due to resonance, the lone pair of electrons on the oxygen atom of phenol is delocalised into the benzene ring, so it is not easily available for protonation.

Teacher's Note:
a) The key word is "resonance" or "delocalisation of lone pair".
b) The same effect makes the oxygen of phenol partially positive.

(c) Which is a stronger acid - phenol or cresol ? Give reason. [1 Mark]

Answer: Phenol is a stronger acid. In cresol, the electron releasing (+I) effect of the methyl group makes the phenoxide ion (cresoxide ion) less stable.

Teacher's Note:
a) Naming phenol earns half a mark and the reason earns the other half.
b) Electron releasing groups decrease acidity; electron withdrawing groups (like -NO₂) increase it.

OR

(c) Write the IUPAC name of the product formed in the Reimer-Tiemann reaction. [1 Mark]

Answer: 2-Hydroxybenzaldehyde (2-hydroxybenzenecarbaldehyde), also called salicylaldehyde.

Teacher's Note:
a) Reimer-Tiemann reaction uses CHCl₃ and aqueous NaOH, followed by acid, and puts -CHO ortho to -OH.
b) Give the IUPAC name, not only the common name.

SECTION E

31. (A) (a) Carry out the following conversions :
(i) Ethanal to But-2-enal
(ii) Propanoic acid to ethane
(b) An alkene A with molecular formula C₅H₁₀ on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C. [5 Marks]

Answer:
(a) (i) Aldol condensation followed by dehydration:
2CH₃CHO →(dil. NaOH) CH₃-CH(OH)-CH₂-CHO (3-hydroxybutanal) →(heat, -H₂O) CH₃-CH=CH-CHO (but-2-enal)
(ii) Sodium salt formation and decarboxylation with soda lime:
CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O
CH₃CH₂COONa + NaOH →(CaO, heat) CH₃-CH₃ + Na₂CO₃
(b) A = (CH₃)₂C=CHCH₃ (2-methylbut-2-ene)
B = CH₃CHO (ethanal): an aldehyde, so it gives Fehling test; it has a CH₃CO- group, so it gives iodoform test.
C = CH₃COCH₃ (acetone / propanone): a ketone, so no Fehling test, but it forms iodoform.
Ozonolysis: (CH₃)₂C=CHCH₃ →(i) O₃, (ii) Zn/H₂O CH₃COCH₃ + CH₃CHO

Teacher's Note:
a) Each conversion carries 1 mark and each of A, B and C carries 1 mark.
b) Fehling test is given by aliphatic aldehydes only; iodoform test needs a CH₃CO- group (or CH₃CH(OH)- group).
c) Check the formula: the carbons of B (2) and C (3) must add up to the 5 carbons of A.

OR

(B) An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to get a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives But-l-ene. Identify (A), (B) and (C) and write chemical equations for the reactions involved. [5 Marks]

Answer:
A = C₃H₇COOC₄H₉ (butyl butanoate)
B = C₃H₇COOH (butanoic acid)
C = C₄H₉OH (butan-1-ol)
Reactions:
1. Hydrolysis: CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃ + H₂O →(dil. H₂SO₄) CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH
2. Dehydration: CH₃CH₂CH₂CH₂OH →(conc. H₂SO₄, heat) CH₃CH₂CH=CH₂ + H₂O
3. Oxidation: CH₃CH₂CH₂CH₂OH →(CrO₃/CH₃COOH, chromic acid) CH₃CH₂CH₂COOH

Teacher's Note:
a) Since oxidation of C gives B, the acid and the alcohol have the same carbon chain (4 carbons each).
b) Check: C₃H₇COOC₄H₉ has the formula C₈H₁₆O₂.
c) But-1-ene from dehydration shows that C is the straight chain primary alcohol, butan-1-ol.

32. (A) In the following complex ions, explain the type of hybridisation, shape and magnetic property : [5 Marks]
(a) [Fe(H₂O)₆]²⁺
(b) [NiCl₄]^2-
(At. Nos. : Fe = 26, Ni = 28)

Answer:
(a) [Fe(H₂O)₆]²⁺:
1. Fe²⁺ has the configuration 3d⁶. H₂O is a weak field ligand, so no pairing occurs; the 3d orbitals have 1 paired and 4 unpaired electrons.
2. The six H₂O ligands use one 4s, three 4p and two 4d orbitals (outer orbital complex).
3. Hybridisation: sp³d²; shape: octahedral; magnetic property: paramagnetic (4 unpaired electrons).
(b) [NiCl₄]^2-:
4. Ni²⁺ has the configuration 3d⁸. Cl^- is a weak field ligand, so the two unpaired electrons in 3d stay unpaired.
5. The four Cl^- ligands use one 4s and three 4p orbitals. Hybridisation: sp³; shape: tetrahedral; magnetic property: paramagnetic (2 unpaired electrons).

Teacher's Note:
a) Show the orbital box diagram for each ion; it carries 1 mark each.
b) Hybridisation, shape and magnetic property carry half a mark each.
c) Weak field ligands (H₂O, Cl^-) do not force pairing, so the complexes are high spin.

OR

(B) (a) Write IUPAC names of the following :
(i) [Co(H₂O)(CN)(en)₂]²⁺
(ii) [PtCl₄]^2-
(iii) [Cr(NH₃)₄Cl(ONO)]⁺
(b) What is spectrochemical series ? Write the difference between a strong field ligand and a weak field ligand. [5 Marks]

Answer:
(a) (i) Aquacyanidobis(ethane-1,2-diamine)cobalt(III) ion
(ii) Tetrachloridoplatinate(II) ion
(iii) Tetraamminechloridonitrito-O-chromium(III) ion
(b) 1. Spectrochemical series: the arrangement of ligands in increasing order of their crystal field splitting (field strength). It is an experimentally determined series based on the absorption of light by complexes with different ligands.
2. For a weak field ligand, \( \Delta_o \lt P \) and it forms high spin complexes; for a strong field ligand, \( \Delta_o \gt P \) and it forms low spin complexes.

Teacher's Note:
a) Ligands are named in alphabetical order (aqua, cyanido, ethane-1,2-diamine), and "bis" is used for the bidentate ligand en.
b) An anionic complex ends in "-ate", as in tetrachloridoplatinate(II).
c) Here P is the pairing energy and \( \Delta_o \) is the crystal field splitting energy.

33. (A) (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K :
Sn(s) | Sn²⁺ (0.004 M) || H⁺ (0.02 M) | H₂(g) (1 Bar) | Pt (s)
(Given : \( E^{\circ}_{Sn^{2+}/Sn} = -0.14 \, V \), \( E^{\circ}_{H^{+}|H_2(g), Pt} = 0.00 \, V \))
(b) Account for the following ;
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH₃COOH decreases on dilution. [5 Marks]

Answer:
(a) Cell reaction: Sn(s) + 2H⁺(aq) → Sn²⁺(aq) + H₂(g); n = 2
Nernst equation: \( E_{cell} = (E^{\circ}_{c} - E^{\circ}_{a}) - \dfrac{0.059}{2} \log \dfrac{[Sn^{2+}]}{[H^{+}]^2} \)
\( E_{cell} = [0 - (-0.14)] - \dfrac{0.059}{2} \log \dfrac{0.004}{(0.02)^2} \)
\( E_{cell} = 0.14 - 0.0295 \log \dfrac{0.004}{0.0004} = 0.14 - 0.0295 \log 10 \)
\( E_{cell} = 0.14 - 0.0295 = 0.1105 \, V \)
(b) (i) Oxygen has a high overpotential (overvoltage) at the anode, so its evolution is kinetically slow and Cl₂ is liberated instead.
(ii) On dilution, the number of ions carrying current per unit volume decreases, so the conductivity of CH₃COOH decreases.

Teacher's Note:
a) The cell reaction, the Nernst equation and the final value carry 1 mark each.
b) Square the [H⁺] term because 2 H⁺ ions take part in the reaction.
c) Do not confuse conductivity (decreases on dilution) with molar conductivity (increases on dilution).

OR

(B) (a) Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use.
(b) Calculate the potential for half-cell containing 0.01 M K₂Cr₂O₇(aq), 0.01 M Cr³⁺ (aq) and 1.0 × 10^-4 M H⁺(aq).
The half cell reaction is
Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- → 2Cr³⁺(aq) + 7H₂O(l)
and the standard electrode potential is given as E° = 1.33 V.
[Given : log 10 = 1] [5 Marks]

Answer:
(a) At anode: Pb(s) + SO₄^2-(aq) → PbSO₄(s) + 2e^-
At cathode: PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) + 2e^- → PbSO₄(s) + 2H₂O(l)
Overall reaction: Pb(s) + PbO₂(s) + 2SO₄^2-(aq) + 4H⁺(aq) → 2PbSO₄(s) + 2H₂O(l)
(b) Given: [Cr₂O₇^2-] = 0.01 M, [Cr³⁺] = 0.01 M, [H⁺] = \( 1.0 \times 10^{-4} \) M, n = 6, E° = 1.33 V
Nernst equation: \( E = E^{\circ} - \dfrac{0.059}{n} \log \dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^{+}]^{14}} \)
\( E = 1.33 - \dfrac{0.059}{6} \log \dfrac{(10^{-2})^2}{(10^{-2})(10^{-4})^{14}} \)
\( E = 1.33 - \dfrac{0.059}{6} \log 10^{54} = 1.33 - \dfrac{0.059}{6} \times 54 \)
\( E = 1.33 - 0.059 \times 9 = 1.33 - 0.531 \)
\( E = 0.799 \, V \)

Teacher's Note:
a) In the Nernst equation for a reduction half-cell, the product terms go in the numerator and the reactant terms in the denominator.
b) Raise [H⁺] to the power 14 and [Cr³⁺] to the power 2, as in the balanced equation.
c) Water is a pure liquid, so it is not included in the log term.

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